The mass per unit length of a rod of length 3 | vedclass

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(B) Let the mass per unit length be $\lambda = kx$,where $k$ is a constant.Consider a small element of length $dx$ at a distance $x$ from the end $x=0

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$2$

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(B) Let the mass per unit length be $\lambda = kx$,where $k$ is a constant.Consider a small element of length $dx$ at a distance $x$ from the end $x=0$.The mass of this element is $dm = \lambda dx = kx dx$.The center of mass $x_{cm}$ is given by the formula:$x_{cm} = \frac{\int x dm}{\int dm}$Substituting the values:$x_{cm} = \frac{\int_0^3 x (kx dx)}{\int_0^3 kx dx} = \frac{\int_0^3 x^2 dx}{\int_0^3 x dx}$Evaluating the integrals:$x_{cm} = \frac{[x^3/3]_0^3}{[x^2/2]_0^3} = \frac{27/3}{9/2} = \frac{9}{4.5} = 2 \ m$.

The mass per unit length of a rod of length $3 \ m$ varies in proportion to the distance $x$ from one of its ends. The center of mass of this rod from that end is at a distance of ........ $m$.

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